Discussion
Fourier sine transform of
\[ f(t) = \frac{e^{-at}}{t},\ a \gt 0 \]
is :
Answer: A
The Fourier sine transform is \(\sqrt{2/\pi}\,\tan^{-1}(s/a)\).
Let \(I(s) = \int_{0}^{\infty} \frac{e^{-at}}{t}\sin st\,dt\). Differentiating under the integral sign,
\[ \begin{aligned} I'(s) &= \int_{0}^{\infty} e^{-at}\cos st\,dt \\ &= \frac{a}{a^{2} + s^{2}} \end{aligned} \]
With \(I(0) = 0\), integrating gives \(I(s) = \tan^{-1}(s/a)\). With the factor \(\sqrt{2/\pi}\) of the sine transform, the answer is \(\sqrt{2/\pi}\tan^{-1}(s/a)\).
The other options cannot be right: a sine transform of this decaying function is bounded and increases to \(\sqrt{2/\pi}\cdot\pi/2\), while \(\tan(s/a)\) blows up and \(\sin^{-1}(s/a)\) is undefined for \(s \gt a\).
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